`int(dx)/(sqrt(4x^(2)+1))`

Answer

To solve the integral \(\int \frac{dx}{\sqrt{4x^2 + 1}}\), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{dx}{\sqrt{4x^2 + 1}} \]

### Step 2: Simplify the Expression Inside the Square Root
Notice that \(4x^2 + 1\) can be rewritten as:
\[ \sqrt{4x^2 + 1} = \sqrt{(2x)^2 + 1} \]
This allows us to use a trigonometric substitution.

### Step 3: Use Substitution
Let \(2x = t\). Then, we have:
\[ dx = \frac{dt}{2} \]
Now substitute \(2x\) and \(dx\) into the integral:
\[ \int \frac{dx}{\sqrt{4x^2 + 1}} = \int \frac{\frac{dt}{2}}{\sqrt{t^2 + 1}} = \frac{1}{2} \int \frac{dt}{\sqrt{t^2 + 1}} \]

### Step 4: Integrate
The integral \(\int \frac{dt}{\sqrt{t^2 + 1}}\) is a standard integral, which equals \(\ln(t + \sqrt{t^2 + 1}) + C\). Therefore, we have:
\[ \frac{1}{2} \int \frac{dt}{\sqrt{t^2 + 1}} = \frac{1}{2} \ln(t + \sqrt{t^2 + 1}) + C \]

### Step 5: Substitute Back
Now, substitute \(t = 2x\) back into the expression:
\[ = \frac{1}{2} \ln(2x + \sqrt{(2x)^2 + 1}) + C \]
This simplifies to:
\[ = \frac{1}{2} \ln(2x + \sqrt{4x^2 + 1}) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{dx}{\sqrt{4x^2 + 1}} = \frac{1}{2} \ln(2x + \sqrt{4x^2 + 1}) + C \]

---